递归与二叉树_leetcode404

# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None


# 思考误区:当只有根节点时,这个根是不能同时作为子节点的,当时在这一点上,思考不过来

class Solution(object):
def sumOfLeftLeaves(self, root):
"""
:type root: TreeNode
:rtype: int
"""

if not root:
return 0

if root.left and not root.left.left and not root.left.right:

return root.left.val + self.sumOfLeftLeaves(root.right)


return self.sumOfLeftLeaves(root.left)+self.sumOfLeftLeaves(root.right)







原文地址:https://www.cnblogs.com/lux-ace/p/10546814.html